$\lim_{x\to -4}\dfrac{x+4}{\sqrt{3x+13}-1}=$
Explanation: Substituting $x=-4$ into $\dfrac{x+4}{\sqrt{3x+13}-1}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{x+4}{\sqrt{3x+13}-1} \\\\ &=\dfrac{x+4}{\sqrt{3x+13}-1}\cdot\dfrac{\sqrt{3x+13}+1}{\sqrt{3x+13}+1} \gray{\text{Rationalize the denominator}} \\\\ &=\dfrac{(x+4)(\sqrt{3x+13}+1)}{(3x+13)-1^2} \\\\ &=\dfrac{\cancel{(x+4)}(\sqrt{3x+13}+1)}{3\cancel{(x+4)}} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{\sqrt{3x+13}+1}{3} \text{, for }x\neq -4 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $-4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{x+4}{\sqrt{3x+13}-1}=\dfrac{\sqrt{3x+13}+1}{3}$ for all $x$ -values in the interval $(-4.5,-3.5)$ except for $x=-4$. Therefore, $\lim_{x\to -4}\dfrac{x+4}{\sqrt{3x+13}-1}=\lim_{x\to -4}\dfrac{\sqrt{3x+13}+1}{3}=\dfrac{2}{3}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to -4}\dfrac{x+4}{\sqrt{3x+13}-1}=\dfrac{2}{3}$.